Correcting The Correction

Maybe,

$\cfrac{q^2}{4\pi \varepsilon_o r^2}=\cfrac{q}{4\pi \sqrt{\varepsilon_o} r^2}.\cfrac{q}{\sqrt{\varepsilon_o}}$

that, the change in $KE$ per unit test charge is,

$\cfrac{q}{4\pi \sqrt{\varepsilon_o} r^2}$

This would change the definition of $\varepsilon_o$ in this model, and the symbol $\sqrt{\varepsilon}$ replaces $\varepsilon_o$. The correction factor to $c$ considering $\sqrt{\varepsilon}$ remains,

$\cfrac {2ln(cosh(\theta_{\psi})) }{ \mu _{ o }  }$

the same.